O-Level Additional Math

[emerald]

Hi, pls help to solve:


(2h - 7k)(3k - 1)(3 - h)


Thanks.

[iFruit]

emerald:

Hi, pls help to solve:


(2h - 7k)(3k - 1)(3 - h)


Thanks.

Hi emerald,

Is the question complete? Could you clarify?

[emerald]

iFruit:

emerald:

Hi, pls help to solve:


(2h - 7k)(3k - 1)(3 - h)


Thanks.

Hi emerald,

Is the question complete? Could you clarify?

Hi iFruit,

This is the given question but maybe there's some problem with it cos my dd couldn't solve it. Anyway, thanks for responding.

[OK Lor]

Hi,


The question is a little off topic, please help to factorise x⁴+ 4

Thanks.

[iFruit]

OK Lor:

Hi,


The question is a little off topic, please help to factorise x⁴+ 4

Thanks.

x⁴+ 4 = x⁴+ 4+4x²-4x² = (x²+2)² - (2x)² = (x²+2x+2)(x²-2x+2)

[OK Lor]

Hi iFruit,


Thank you very much

[atutor2001]

iFruit:

OK Lor:

Hi,


The question is a little off topic, please help to factorise x⁴+ 4

Thanks.

x⁴+ 4 = x⁴+ 4+4x²-4x² = (x²+2)² - (2x)² = (x²+2x+2)(x²-2x+2)

Hi iFruit

I really enjoy your solution, it appears so simple once the approach is correct.

Just curious, is the above approach covered in normal O level course work for A math (it appears as a modification to \"completing the square\" to me) or is it something that we gain through more exposure to mathematical problems.

Regards

[iFruit]

atutor2001:

iFruit:

[quote=\"OK Lor\"]Hi,


The question is a little off topic, please help to factorise x⁴+ 4

Thanks.

x⁴+ 4 = x⁴+ 4+4x²-4x² = (x²+2)² - (2x)² = (x²+2x+2)(x²-2x+2)

Hi iFruit

I really enjoy your solution, it appears so simple once the approach is correct.

Just curious, is the above approach covered in normal O level course work for A math (it appears as a modification to \"completing the square\" to me) or is it something that we gain through more exposure to mathematical problems.

Regards[/quote]Hi atutor2001,

You are right. IMHO, as kids gain experience they can routinely solve these kind of problems. Just a matter of exposure..

Regards.

[small]

Hi all,


Please help to solve this Sec 1 question and many thanks in advance.

http://s1.postimage.org/sxVwS-0c78d7a2.jpg\">


.

[iFruit]

small:

Hi all,


Please help to solve this Sec 1 question and many thanks in advance.

http://s1.postimage.org/sxVwS-0c78d7a2.jpg\">


.

Pi = π

The three vertices (lets say A, B, C) forming the shaded area are the centers of the three circles.

So if you join the three vertices by straight lines, it will be an equilateral triangle of radius 1cm

The area of the segment BC = Area of sector ABC - Area of triangle ABC
= 1/6 (area of circle) - 1/2 (base)(height)

=π/6 -1/2 x 1x √(1-1/4) = π/6 - (√3)/4

So area of the shaded part = 3x Segment BC + Area of triangle

= 3(π/6 - (√3)/4) + (√3)/4 = π/2 -(√3)/2 = 0.5(π-√3)


HTH